In a hypothetical fission reaction 92 X 236 rightarrow 56 Y 141 + 36 Z 92 +3 R The identity of emitted particles (R) is :

Physics · Nuclear Physics

JEE Main 2024 — 8 April, Shift 2 — Question 30

In a hypothetical fission reaction ${ }_{92} \mathrm{X}^{236} \rightarrow{ }_{56} \mathrm{Y}^{141}+{ }_{36} \mathrm{Z}^{92}+3 \mathrm{R}$ The identity of emitted particles $(R)$ is :

Option A:
Proton
Option B:
Electron
Option C:
Neutron
Correct
Option D:
$\gamma$ -radiations

Answer: C

Step-by-step solution

Z in LHS $=92$

Z in $\mathrm{RHS}=56+36=92$

A in LHS $=236$

A in RHS $=141+92=233$

So 3 neutrons are released.

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 8 April, Shift 2
Subject: Physics
Chapter: Nuclear Physics
Topic: Nuclear Fission and Fusion

If M o is the mass of isotope 5^12 B, M p and M n are the masses of proton and neutron, then nuclear binding energy of isotope is :

Answer: C

Step-by-step solution

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